The sum of the squares of the first n positive integers is always a composite number, except the case of n=2.
Prove it.
The formula for the sum is S=n*(n+1)*(2n+1)/6. The first few sums are 1, 5, 14, 30, 55.
At least one of the three terms in the product is even; exactly one is a multiple of 3. If the same term is both even and a multiple of 3 then the product of the remaining two terms is composite for any n>=2 (n=5 is the first n which qualifies in this case).
If the multiple of 3 is odd, then dividing an even term by 2 will be greater than 1 for n>=3; multiplying that term by the other non-multiple of 3 will be composite.
The combination of both cases cover all n>=3. The only noncomposite values the sum can be is 1 or 5, corresponding to n=1 or 2.
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